Answer
(a) The ball's speed at the lowest point is 1.4 m/s
(b) The pendulum will swing up to an angle of $30^{\circ}$ on the other side.
Work Step by Step
(a) Let $L$ be the length of the pendulum. We can find the height $h$ (above the ball's lowest point) when the pendulum is at a $30^{\circ}$ angle.
$\frac{L-h}{L} = cos(\theta)$
$h = L~[1-cos(\theta)]$
$h = (0.75~m)~[1-cos(30^{\circ})]$
$h = 0.10~m$
We can use conservation of energy to find the ball's speed at the lowest point. The kinetic energy at the lowest point will be equal to the potential energy at height $h$. So;
$KE = PE$
$\frac{1}{2}mv^2 = mgh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(0.10~m)}$
$v = 1.4~m/s$
The ball's speed at the lowest point is 1.4 m/s
(b) By conservation of energy, the ball will go up to the same height $h = 0.10~m$ on the other side. Therefore, the pendulum will swing up to an angle of $30^{\circ}$ on the other side.
